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Take $X=[1,2]\cup [3,4]\cup [5,6]\cup\dots$. Now take the quotient topology by identifying the points $1,3,5,\dots$ (the left end of each interval in the union). My book says that this space $X/\sim$ is not first countable.

I wonder why that is. Consider the countable collection of intervals $\{ [2k+1,2k+1+\frac{1}{n})\}_{n=1}^\infty$ for each $k\in\Bbb{N}\cup\{0\}$.

Doesn't this prove that $X/\sim$ is first countable?

  • Well, is $[2k + 1, 2k + 1 + \frac{1}{n})$ open in the quotient topology? And if so, given $x \in X$, and $U$ an open neighborhood of $x$, can we find some $[2k+ 1, 2k + 1 + \frac{1}{n})$ such that $x \in [2k + 1, 2k+1+ \frac{1}{n}) \subseteq U$? – layman Sep 29 '15 at 17:35
  • What do you wish to do with that collection of intervals? So far, it's just a collection of intervals, and a collection of intervals does not prove anything. – Lee Mosher Sep 29 '15 at 17:57
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    This quotient space is constructed in a similar way as $\Bbb R/\Bbb Z$, where you identify all integers in the real line. Brian's answer in the linked topic can be applied here after minor adjustments. – Stefan Hamcke Sep 29 '15 at 18:06

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The point is, if we pick any sequence $r(k) \in (0,1)$, and define the set $U((r_k)_k) = q[\cup_{k} [2k, 2k+r(k))]$ (where $q$ is the quotient map), then all of these sets are open, and form a local base at the class of integers in $X / \sim$. A diagonal style argument like in the very comparable answer by Brian will show that no countable base can suffice.

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Henno Brandsma
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